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Math Help - Binomial distribution.

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    Binomial distribution.

    65% of the dishwashers manufactured by a factory do not need repairs for the first three years.If 8 of the dishwashers are to be selected by chance, find the probability
    i) all 8 will not need repairs for the first three years
    ii) at least two will need repairs within three years

    someone can help me solve this question?
    Last edited by mr fantastic; July 6th 2011 at 03:41 PM. Reason: Re-titled.
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    Re: Probability Question

    For part i.)

    You know that 65% of dishwashers do NOT need repairs for 3 years. You select 8 dishwashers at random. Each has 65% chance of not needing repairs. We can extrapolate then. The chance of 1 dishwasher not needing repairs = (0.65). The chance of 2 not needing repairs is (0.65)(0.65). You can continue on from there. Does that make sense?
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    Re: Binomial distribution.

    Quote Originally Posted by JWEI View Post
    65% of the dishwashers manufactured by a factory do not need repairs for the first three years.If 8 of the dishwashers are to be selected by chance, find the probability
    i) all 8 will not need repairs for the first three years
    ii) at least two will need repairs within three years

    someone can help me solve this question?
    I re-titled your post "Binomial distribution". I assume the question has come from that part of your work? So, define the random variable. Now, what is p, what is n? What calculation is required? Note that Pr(at least 2) = 1 - Pr(less than 2) ....
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    Re: Binomial distribution.

    Make p is the probability of having a dishwasher that does not need repair. Find where k=8 for

    \displaystyle P(k) = ^nC_k \times p^k(1-p)^{n-k}
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  5. #5
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    Re: Binomial distribution.

    Quote Originally Posted by pickslides View Post
    Make p is the probability of having a dishwasher that does not need repair. Find where k=8 for
    \displaystyle P(k) = ^nC_k \times p^k(1-p)^{n-k}
    May I suggest that the following LaTeX notation is easier to follow.
    [tex]P(k)=\binom{n}{k}p^k(1-p)^{n-k}[/tex] gives
     P(k)=\binom{n}{k}p^k(1-p)^{n-k}
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